\(\displaystyle \sum _{k=1}^{10} \frac{k \times 2^k}{(k+1)(k+2)} \)


\(\displaystyle =\sum _{k=1}^{10} k \times 2^k \times (\frac{1}{k+1}-\frac{1}{k+2}) \)


\(\displaystyle =1+\sum _{k=1}^9 \frac{1}{k+2}[(k+1) \times 2^{k+1}-k \times 2^k]-\frac{10 \times 2^{10}}{12} \)


\(\displaystyle =1+\sum _{k=1}^9 \frac{1}{k+2} \times 2^k \times (2k+2-k)-\frac{5 \times 2^9}{3} \)


\(\displaystyle =1+2+4+ \cdots +512-\frac{2560}{3} \)


\(\displaystyle =\frac{509}{3} \)

你把前幾項寫出來，然後分母相同的收在一起就可以了。

瑋岳老師居然用這麼神奇的招式~~
\(\ P(6,-2),A(a,0),B(0,b),Q(-4,3) \)
所求為\( PA+AB+BQ \)的最小值